Difference between revisions of "2005 AIME II Problems/Problem 3"
Matrix math (talk | contribs) m (→Solution 2) |
Mathgenius (talk | contribs) m (→Solution 1) |
||
Line 3: | Line 3: | ||
== Solution 1 == | == Solution 1 == | ||
− | Let's call the first term of the original [[geometric series]] <math>a</math> and the common ratio <math>r</math>, so <math>2005 = a + ar + ar^2 + \ldots</math>. Using the sum formula for [[infinite]] geometric series, we have <math> | + | Let's call the first term of the original [[geometric series]] <math>a</math> and the common ratio <math>r</math>, so <math>2005 = a + ar + ar^2 + \ldots</math>. Using the sum formula for [[infinite]] geometric series, we have <math>\;\;\frac a{1 -r} = 2005</math>. Then we form a new series, <math>a^2 + a^2 r^2 + a^2 r^4 + \ldots</math>. We know this series has sum <math>20050 = \frac{a^2}{1 - r^2}</math>. Dividing this equation by <math>\frac{a}{1-r}</math>, we get <math>10 = \frac a{1 + r}</math>. Then <math>a = 2005 - 2005r</math> and <math>a = 10 + 10r</math> so <math>2005 - 2005r = 10 + 10r</math>, <math>1995 = 2015r</math> and finally <math>r = \frac{1995}{2015} = \frac{399}{403}</math>, so the answer is <math>399 + 403 = \boxed{802}</math>. |
(We know this last fraction is fully reduced by the [[Euclidean algorithm]] -- because <math>4 = 403 - 399</math>, <math>\gcd(403, 399) | 4</math>. But 403 is [[odd integer | odd]], so <math>\gcd(403, 399) = 1</math>.) | (We know this last fraction is fully reduced by the [[Euclidean algorithm]] -- because <math>4 = 403 - 399</math>, <math>\gcd(403, 399) | 4</math>. But 403 is [[odd integer | odd]], so <math>\gcd(403, 399) = 1</math>.) |
Revision as of 20:02, 7 March 2019
Contents
Problem
An infinite geometric series has sum 2005. A new series, obtained by squaring each term of the original series, has 10 times the sum of the original series. The common ratio of the original series is where and are relatively prime integers. Find
Solution 1
Let's call the first term of the original geometric series and the common ratio , so . Using the sum formula for infinite geometric series, we have . Then we form a new series, . We know this series has sum . Dividing this equation by , we get . Then and so , and finally , so the answer is .
(We know this last fraction is fully reduced by the Euclidean algorithm -- because , . But 403 is odd, so .)
Solution 2
We can write the sum of the original series as , where the common ratio is equal to . We can also write the sum of the second series as . Using the formula for the sum of an infinite geometric series , where is the sum of the sequence, is the first term of the sequence, and is the ratio of the sequence, the sum of the original series can be written as , and the second sequence can be written as . Dividing by , we obtain , which can also be written as . Substitute this value for back into , we obtain . Dividing both sides by 10 yields we can now write a system of equations with and , but this does not output integer solutions. However, we can also write as . This gives the system of equations and , which does have integer solutions. Our answer is therefore (Solving for and gives us and , respectively, which are co-prime).
See also
2005 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.